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 A165783 a(n) = A002326(n-1) + A000120(A165781(n-1)). 1
 2, 3, 6, 4, 9, 15, 18, 5, 12, 27, 8, 15, 30, 27, 42, 6, 15, 17, 54, 16, 30, 21, 17, 32, 31, 10, 78, 28, 27, 87, 90, 7, 18, 99, 33, 49, 12, 29, 45, 56, 81, 123, 10, 39, 15, 16, 13, 50, 72, 45, 150, 74, 16, 159, 54, 50, 42, 63, 15, 33, 165, 26, 150, 8, 21, 195, 26, 53, 102, 207 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Given a shift register : r(k)=r(k-1)+ X if r(k-1) is not divisible Y, else r(k)=r(k-1)/Y. Gcd(r(0), X))=1, Gcd(X, Y)=1. Then the length of the period orbit of such a register is L + digitsum (r(L)*(Y^L-1)/ X). Digitsum(z)in base X. r(L) a point from period orbit, L minimal possible exponent such that (Y^L-1)/X)is a positive integer. Number of period orbits is the order of the cyclic group connected to the register. a(n) is the period length for Y=2, X=2*n-1, r(L)=1. [Ctibor O. Zizka, Nov 24 2009] LINKS Ivan Neretin, Table of n, a(n) for n = 1..10000 FORMULA a(n) = L + digitsum((2^L -1)/(2*n-1)). Digitsum(z)in base 2. [Ctibor O. Zizka, Nov 24 2009] EXAMPLE n=1, a(1)=1 + digitsum(1)= 2. n=2, a(2)=2 + digitsum(1)=3. n=3, a(3)= 4 + digitsum(3) = 6. n=4, a(4)= 3 + digitsum(1)=4. n=5, a(5)= 6 + digitsum(7)=9. [Ctibor O. Zizka, Nov 24 2009] MAPLE A002326 := proc(n) if n = 0 then 1; else numtheory[order](2, 2*n+1) ; end if ; end proc: A165781 := proc(n) (2^A002326(n)-1)/(2*n+1) ; end proc: read("transforms") ; A165783 := proc(n) A002326(n-1)+wt(A165781(n-1) ) ; end proc: seq(A165783(n), n=1..80) ; # R. J. Mathar, Nov 26 2009 MATHEMATICA Table[(b = MultiplicativeOrder[2, 2 n - 1]) + Plus @@ IntegerDigits[(2^b - 1)/(2 n - 1), 2], {n, 1, 70}] (* Ivan Neretin, May 09 2015 *) PROG (PARI) hamming(n)=my(v=binary(n)); sum(i=1, #v, v[i]) a(n)=my(x=2*n+1, m=znorder(Mod(2, x))); m+hamming((1<

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Last modified November 27 06:50 EST 2022. Contains 358362 sequences. (Running on oeis4.)